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The uncertainty in a particle's momentum in an infinite well in the general case of arbitrary nis given bynhL .

Short Answer

Expert verified

For n=0the uncertainty vanishes but is perfectly finite for all other values. Thus, so long asx2 is large enough (it is), the uncertainty principle is perfectly satisfied for all n>0.

Step by step solution

01

The concept and the formula used.

Heisenberg's uncertainty principle states that it is impossible to measure or calculate exactly, both the position and the momentum of an object.

Consider, energy E=0. Then, the momentum of the state must satisfyE=p22m. Now, there are two solutions forP corresponding to positive and negative momentum. The uncertainty can thus be calculated as follows:

螖笔2=P2P2

=2mE

=n222L2

螖笔=苍蟺L

02

Conclusion

Clearly, forn=0 the uncertainty vanishes but is perfectly finite for all other values. Thus, so long asx2 is large enough (it is), the uncertainty principle is perfectly satisfied for alln>0 .

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Most popular questions from this chapter

Write out the total wave function(x,t).For an electron in the n=3 state of a 10nm wide infinite well. Other than the symbols a and t, the function should include only numerical values?

Quantum-mechanical stationary states are of the general form (x,t)=(x)e-it. For the basic plane wave (Chapter 4), this is (x,t)=Aeikxe-it=Aei(kx-t), and for a particle in a box it is Asinkxe-i蝇t. Although both are sinusoidal, we claim that the plane wave alone is the prototype function whose momentum is pure-a well-defined value in one direction. Reinforcing the claim is the fact that the plane wave alone lacks features that we expect to see only when, effectively, waves are moving in both directions. What features are these, and, considering the probability densities, are they indeed present for a particle in a box and absent for a plane wave?

The term interaction is sometimes used interchangeably with force, and other times interchangeably with potential energy. Although force and potential energy certainly aren't the same thing, what justification is there for using the same term to cover both?

Sketch the wave function. Is it smooth?

(x)={2a3xe-axX>00X<0

It is possible to take the finite well wave functions further than (21) without approximation, eliminating all but one normalization constant C . First, use the continuity/smoothness conditions to eliminate A, B , andG in favor of Cin (21). Then make the change of variables z=x-L/2 and use the trigonometric relations

sin(a+b)=sinacosb+cosasinband

cos(a+b)=cosacosb-sinasinbon the

functions in region I, -L/2<z<L/2. The change of variables shifts the problem so that it is symmetric about z=0, which requires that the probability density be symmetric and thus that (z)be either an odd or even function of z. By comparing the region II and region III functions, argue that this in turn demands that (/k)sinkL+coskL must be either +1 (even) or -1 (odd). Next, show that these conditions can be expressed, respectively, as k=tankL2 and k=-cotkL2. Finally, plug these separately back into the region I solutions and show that

(z)=C{e(z+L/2)鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌z<L/2coskzcoskL2鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌-L/2<z<L/2e-(z-L/2)鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌z>L/2


or

(z)=C{e(z+L/2)鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌z<L/2-sinkzsinkL2鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌-L/2<z<L/2e-(z-L/2)鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌z>L/2

Note that Cis now a standard multiplicative normalization constant. Setting the integral of |(z)|2 over all space to 1 would give it in terms of kand , but because we can鈥檛 solve (22) exactly for k(or E), neither can we obtain an exact value for C.

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